How to prove $E(\sum\limits_{n=1}^\infty X_n)=\sum\limits_{n=1}^\infty EX_n$ How to prove that if $X_n>0$, then $E(\sum\limits_{n=1}^\infty X_n)=\sum\limits_{n=1}^\infty EX_n$?
I think I should use something like monotone convergence theorem, but I really don't know how to apply it. Can anyone be kind enough to give me the answer. I will be very grateful!!
 A: All you need is the monotone convergence theorem.
Note that if you write $\mathbb E(\cdot)$ as $\int (\cdot) d\mathbb P $ and also write $\sum_{i=1}^\infty(\cdot)$ as $\lim_{n\to\infty}\sum_{i=1}^n(\cdot)$, you can rewrite your desired statement as
$$\int \lim_{n\to\infty} \sum_{i=1}^n X_i \ d\mathbb P = \lim_{n\to\infty} \int \sum_{i=1}^n X_i \ d\mathbb P$$
And now we define $S_n:= \sum_{i=1}^n X_i$ which is again a random variable (why?). Moreover $(S_n)_{n\in \mathbb N}$ is monotone (why?).
Do you now see how to apply the monotone convergence theorem?
A: It looks like you are looking for Tonelli's theorem.
A: Define $S_{n}=\sum_{k=1}^{n}X_{k}$ and $S=\sum_{k=1}^{\infty}X_{k}$.
Then $S_{n}\uparrow S$ monotonically and Lebesgues monotone convergence theorem
tells us that $\mathbb{E}S_{n}\uparrow\mathbb{E}S$. 
Here $\mathbb{E}S_{n}=\sum_{k=1}^{n}\mathbb{E}X_{k}$
so that $\mathbb{E}S_{n}\uparrow\sum_{k=1}^{\infty}\mathbb{E}X_{k}$
Combination of these facts give: $$\mathbb{E}\left(\sum_{k=1}^{\infty}X_{k}\right)=\sum_{k=1}^{\infty}\mathbb{E}X_{k}$$
